The Goldstone boson associated with a conserved charge is described by $$ |\pi(\vec p)\rangle\propto \int d^3x~e^{i\vec p\cdot\vec x}j_0(x)|0\rangle. $$ How can I see that this is a state with energy given by $E_0+E(\vec p)$ and momentum $\vec p$, namely a one-particle state? Is it for instance possible to use the Hamilton operator in order to check that the given state has the energy $E_0+E(\vec p)$ (and the momentum operator in order check that the state has momentum $\vec{p}$)?
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1$\begingroup$ Schwartz, your text whose (28.8) you are quoting, has reviewed all in his preceding section. Note $\pi(p)$ is not the canonical momentum, but, instead, $\phi(p)$, the pion field itself, and $J_0(x)\propto F\partial_0 \phi(x)$, and its commutation with H yields a time derivative, so $\sim F \nabla \cdot \vec {\mathbf J}(x)\sim F\nabla^2 \phi(x)$... $\endgroup$– Cosmas ZachosCommented Sep 28, 2022 at 15:57
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$\begingroup$ Thanks for the answer. I have the following derivation: Expand $j_0(x)$ as $$ j_0(x)=\int \frac{d^3k}{(2\pi)^3}f(k)a_{\vec k}^\dagger e^{-\vec k\cdot x}$$ where I didn't write the $a_{\vec k}$ term since they annihilate the vacuum. And using $H\sim\int d^3 k\omega_{\vec k}a_{\vec k}a_{\vec k}^\dagger$ we can arrive at the result. Is my derivation correct? $\endgroup$– Louis ChouCommented Sep 29, 2022 at 13:04
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$\begingroup$ It should be, to the extent the current here is linear in the pion field; it's not a deep point: most QFT texts such as Schwarz's, evidently yours, cover it in the review of current algebra manipulations... $\endgroup$– Cosmas ZachosCommented Sep 29, 2022 at 14:04
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